(M)Backtracking:39. Combination Sum
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这道题意思是给你一个数组,再给你一个目标值target,写出所有和是target的子数组。
这个题一开始求出来的结果是[2,2,3][2,3,2][3,2,2][7],发现有重复的,所以先把数组从小到大排序,选择数字的时候只往右边选,不往回选,这样就不会出现重复数组了。
class Solution {public: vector<vector<int>> combinationSum(vector<int>& candidates, int target) { vector<vector<int>> res; vector<int> r; sort(candidates.begin(), candidates.end()); backtracking(candidates, target, res, r, 0, 0); return res; } void backtracking(vector<int>& candidates, int target, vector<vector<int>>& res, vector<int>& r, int sum, int start) { if(sum == target) { res.push_back(r); return; } for(int i = start; i < candidates.size(); ++i) { if(sum + candidates[i] <= target) { r.push_back(candidates[i]); backtracking(candidates, target, res, r, sum + candidates[i], i); r.erase(--r.end()); } } }};
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